Answer
There is necessary a total of 20.6 mL of that $NaOH$ solution to neutralize the $HBr$ one.
Work Step by Step
1. Write and balance the equation for the reaction between $NaOH$ and $HBr$:
$NaOH(aq) + HBr(aq) --\gt NaBr(aq) + H_2O(l)$
- The equation is already balanced.
2. Identify the conversion factors.
- The concentration of that $NaOH$ solution is equal to 0.150 mole per liter.
$\frac{0.150 \space mole \space NaOH}{1 \space L \space (soln.)}$ and $\frac{1 \space L \space (soln.)}{0.150 \space mole \space NaOH}$
- The concentration of that $HBr$ solution is equal to 0.618 mole per liter.
$\frac{0.618 \space mole \space HBr}{1 \space L \space (soln.)}$ and $\frac{1 \space L \space (soln.)}{0.618 \space mole \space HBr}$
- According to the balance equation, each mole of $NaOH$ reacts with one mole of $HBr$:
$\frac{1 \space mole \space HBr}{1 \space mole \space NaOH}$ and $\frac{1 \space mole \space NaOH}{1 \space mole \space HBr}$
3. Calculate the required volume:
$5.00 \space mL \space (Soln. \space HBr) \times \frac{0.618 \space mole \space HBr}{1 \space L \space (soln. \space HBr)} \times \frac{1 \space mole \space NaOH}{1 \space mole \space HBr} \times \frac{1 \space L \space (soln. \space NaOH)}{0.150 \space mole \space NaOH}= 20.6 \space mL \space (Soln. \space NaOH)$
